The area (in sq. units) of the region describ | vedclass

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(D) To find the area of the region bounded by the parabola $y^2 = 2x$ and the line $y = 4x - 1$,we first find their points of intersection. Substituti

Vedclass · Accepted Answer

$\frac{9}{32}$

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(D) To find the area of the region bounded by the parabola $y^2 = 2x$ and the line $y = 4x - 1$,we first find their points of intersection. Substituting $x = \frac{y^2}{2}$ into the line equation: $y = 4(\frac{y^2}{2}) - 1 \implies y = 2y^2 - 1 \implies 2y^2 - y - 1 = 0$. Factoring gives $(2y + 1)(y - 1) = 0$,so $y = 1$ and $y = -\frac{1}{2}$.The area is given by the integral of the right curve minus the left curve with respect to $y$ from $y = -\frac{1}{2}$ to $y = 1$:$Area = \int_{-\frac{1}{2}}^{1} (x_{Right} - x_{Left}) dy = \int_{-\frac{1}{2}}^{1} (\frac{y+1}{4} - \frac{y^2}{2}) dy$$= [\frac{1}{4}(\frac{y^2}{2} + y) - \frac{y^3}{6}]_{-\frac{1}{2}}^{1}$$= [\frac{1}{4}(\frac{1}{2} + 1) - \frac{1}{6}] - [\frac{1}{4}(\frac{1}{8} - \frac{1}{2}) - \frac{(-1/8)}{6}]$$= [\frac{3}{8} - \frac{1}{6}] - [\frac{1}{4}(-\frac{3}{8}) + \frac{1}{48}]$$= [\frac{9-4}{24}] - [-\frac{3}{32} + \frac{1}{48}] = \frac{5}{24} - [\frac{-9+2}{96}] = \frac{5}{24} + \frac{7}{96} = \frac{20+7}{96} = \frac{27}{96} = \frac{9}{32}$ sq. units.

The area (in sq. units) of the region described by $\{(x, y): y^2 \leq 2x\}$ and $\{y \geq 4x - 1\}$ is

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